can any one tell me how to do question 1, 3,4,5,6,7,8? and when u have answer plz remember to explain ur steps thnx
hello, I could give answer to your first question
when a square is inscribed in a circle, the length of the diagnol of the square =the diameter of the circle
2^(1/2)side of square = diagonal of square
area of square = 1/2 * diagonal * diagonal
S1 = 2 (diagonal of S1 =2)
when a circle is inscribed in a square, length of square = diameter of circle
diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
S2 = 1/2 * 2 = 1
Sn = 2*[(1/2)^(n-1)]
2)Originally Posted by kansai
You know that,
$\displaystyle t_1=-2$
$\displaystyle t_2=1$
And that,
$\displaystyle t_3=2t_2-t_1$
thus,
$\displaystyle t_3=2(1)-(-2)=2+2=4$
And that,
$\displaystyle t_4=2t_3-t_2$
thus,
$\displaystyle t_4=2(4)-1=8-1=7$
And that,
$\displaystyle t_5=2t_4-t_3$
thus,
$\displaystyle t_5=2(7)-1=13$
Hello, kansai!
Here's #5 . . .
5. For the matrices: $\displaystyle A = \begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix}$ and $\displaystyle B = \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$
(a) State the value of: $\displaystyle \text{(i) }a_{12}\qquad\text{(ii) }a_{22}\qquad\text{(iii) }a_{31}$
(b) Calculate: $\displaystyle \text{(i) }A - 3B\qquad\text{(ii) }2A + B\qquad\text{(iii) }A'$
$\displaystyle \text{(a) }\;a_{12}$ means the element in the $\displaystyle 1^{st}$ row and the $\displaystyle 2^{nd}$column. .Hence: $\displaystyle a_{12} = 5$
. . . .Similarly: $\displaystyle a_{22} = 0,\;a_{31} = 2$
The elements of matrix $\displaystyle B$ would have been written: $\displaystyle b_{12},\;b_{22},\;b_{31}$
$\displaystyle \text{(b)(i) }A - 3B \:=\:\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - 3\cdot\begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$
. . . $\displaystyle =$ $\displaystyle \;\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} - \begin{bmatrix}6&0&-3\\3&0&3\\9&-6&0\end{bmatrix} \;=\;\begin{bmatrix}-5&5&3\\0&0&1\\-7&6&-1\end{bmatrix}$
$\displaystyle \text{(b)(ii) }2A + B \;= \;2\begin{bmatrix}1&5&0\\3&0&4\\2&0&-1\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix}$
. . . $\displaystyle =\;\begin{bmatrix}2&10&0\\6&0&8\\4&0&-2\end{bmatrix} + \begin{bmatrix}2&0&-1\\1&0&1\\3&-2&0\end{bmatrix} \;= \;\begin{bmatrix}4&10&-1\\7&0&9\\7&-2&-2\end{bmatrix}$
$\displaystyle \text{(b)(iii) }A'$ . . . I assume this means the inverse of $\displaystyle A.$
The steps are long and messy . . . so I'll omit them.
. . . $\displaystyle A'\;=\;\frac{1}{55}\begin{bmatrix}0&5&20\\11&-1&-4\\0&10&-15\end{bmatrix} $
Hello, kansai!
Here's #3 . . .
For the map given on the right
(a) Represent the map with a network.
(b) Find the degree of each vertex.
(c) State whether the network is traceable and explain the reason.(b) A: degree 2Code:(a) A - - - - - B | / | \ | / | \ | C | \ | / | \ | / | \ D - - - - - E - - -F
. . .B: degree 4
. . .C: degree 2
. . .D: degree 3
. . .E: degree 3
. . .F: degree 2
(c) A network is traceable if it has at most two vertices of odd degree.
This network has two odd vertices (D and E) and hence is traceable.
Hello, kansai!
I don't "trust" #4 . . . It has some disturbing errors and the meaning is unclear.
. . Can you provide the original wording?
A petty point: I assume it refers to The "NYS school system".4. The NYS school has four committees.
Each of these committees meet once a month.
Membership on these committees is as follows:
. . Committee A: Anna, John, Steven, Tom
. . Committee B: Debra, Jane John
. . Committee C: Annie, Debra, Larry, Tom
. . Committee D: Bobby, Debra, Grace, Mary, Steven
. . Committee E: Anna, Larry, Tom.
(a) Draw a network to illustrate the connectivity among the committees. .?
(b) Design a schedule with minimum timeslots for the committee meeting without conflicts. .??
I would assume that Annie is the same person as Anna, that someone mistyped the names.
. . But do I dare make that assumption?
The most glaring errors is that there are five committees.
(a) Connectivity is a term used in network theory, but not like this.
. . I assume they mean "common memberships" ?
(b) Minimum timeslots is a very sloppy term.
Does it mean "the shortest meetings"?
. . How about: "I call the meeting to order. .I will entertain a motion to adjourn . . ."
Does it mean "the shortest time-span" for the four (five?) meetings?
. .Schedule them consecutively with a one-minute break in between
. . (to allow the ten people to switch seats).
It is obvious that no two committees can meet at the same time.
. . So what is the question?
If I am correct about part (a), this should suffice:Code:B / | \ / | \ / | \ A - - - + - - - D / \ | / / \ | / / \ | / E - - - - - C
From my understanding of this question it works out like this.
the diagnol of the square (d) = the diameter of the circle (d), so the area =
$\displaystyle s$ = the length of one of the square's side
$\displaystyle d=\sqrt{ s^2 +s^2} $
$\displaystyle d^2=s^2 +s^2 $
$\displaystyle d^2=2s^2 $
$\displaystyle \frac{d^2}{2}=s^2 $ since s^2 is the area of the square this is the answer for $\displaystyle S_1 $.
Now the next circle has the diameter equal to the height of the square, so...
$\displaystyle \sqrt{\frac{d^2}{2}}=s=d_2 $
and now you apply the first formula to find the area of $\displaystyle S_2 $
$\displaystyle \sqrt{\frac{d^2}{2}}=\sqrt{s^2 +s^2} $
$\displaystyle \sqrt{\left(\frac{d^2}{2}\right)^2}=s^2 +s^2 $
$\displaystyle \frac{d^2}{2}=2s^2 $
$\displaystyle \frac{\frac{d^2}{2}}{2}=s^2 $
$\displaystyle \frac{d^2}{4}=s^2 $
$\displaystyle \frac{d^2}{4}=s^2 $
This shows us that the area of $\displaystyle S_n = \frac{d^2}{2^n} $
Considering me and malaygoel got different answers, could a third party check my work?
it might be that I'm tired, buy I can't seem to think where you got the 2^(-1/2) from. However, you can see our equations differ,diameter of O2= length of S1 = 2^(-1/2)* diagonal of S1 = 2^(1/2)
mine= $\displaystyle \frac{d^2}{2^n} $ yours= $\displaystyle 2\left(\frac{1}{2^{(n-1)}}\right) $
However, there is a certain value of d that would make our equations the same.
$\displaystyle \frac{d^2}{2^n}=2\left(\frac{1}{2^{(n-1)}}\right) $
$\displaystyle \frac{d^2}{2^n}=\frac{2}{2^{(n-1)}} $
$\displaystyle \frac{2^2}{2^n}=\frac{2}{2^{(n-1)}} $
$\displaystyle \frac{4}{2^n}=\frac{2}{2^{(n-1)}} $
$\displaystyle \frac{2}{2^{(n-1)}}=\frac{2}{2^{(n-1)}} $
but why would you say the diameter is 2?
assuming that the diameter of the first circle is 2, then yes.
$\displaystyle \frac{d^2}{2^n} $
$\displaystyle \frac{2^2}{2^n} $
$\displaystyle \frac{4}{2^n} $
$\displaystyle \frac{4}{2^n} $
$\displaystyle \frac{1}{2^{(n-2)}} $
$\displaystyle 2^{(-1)(n-2)} $
$\displaystyle 2^{(-n+2)} $
$\displaystyle 2^{(2-n)} $